Question

A large on-demand, video streaming company is designing a large-scale survey to determine the mean amount of time corporate executives watch on-demand television. A small pilot survey of 10 executives indicated that the mean time per week is 12 hours, with a standard deviation of 3 hours. The estimate of the mean viewing time should be within 0.25 hour. The 95% level of confidence is to be used. How many executives should be surveyed? (Use z Distribution Table.)
How many executives should be surveyed? (Round the final answer to the next whole number.)

Answer 1

Answer:

554 executives should be surveyed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$.

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Standard deviation of 3 hours.

This means that $\sigma = 3$

The 95% level of confidence is to be used. How many executives should be surveyed?

n executives should be surveyed, and n is found with $M = 0.25$. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.25 = 1.96\frac{3}{\sqrt{n}}$

$0.25\sqrt{n} = 1.96*3$

$\sqrt{n} = \frac{1.96*3}{0.25}$

$(\sqrt{n})^2 = (\frac{1.96*3}{0.25})^2$

$n = 553.2$

Rounding up:

554 executives should be surveyed.